# Relation between coefficients and zeros of a quadratic polynomial

1. May 20, 2012

### physics kiddy

1. The problem statement, all variables and given/known data

For any quadratic polynomial ax2+bx+c having zeros β and α
Prove that β + α = -b/a and αβ = c/a.

2. Relevant equations

3. The attempt at a solution

I have found a method myself to prove α+ β = -b/a. However, I could not prove αβ = c/a.

It goes like this.

If α and β are the zeros of the given polynomial.

a(α)2+b(α) + c = 0 .................... (i)

Also,

a(β)2+b(β) + c = 0 ....................(ii)

Comparing (i) and (ii)

a(α)2+b(α) + c = a(β)2+b(β) + c

=> aα2-aβ2 = bβ - bα
=>a(α22) = -b(α-β)
=>α22/α-β = -b/a
=> α+β = -b/a

Please help me prove αβ = c/a using the same method.

Last edited: May 20, 2012
2. May 20, 2012

### SammyS

Staff Emeritus
Use the quadratic formula for the two zeros of the quadratic polynomial.

3. May 20, 2012

### Hurkyl

Staff Emeritus
Do you have reason to think the same method will work?

It would be easier to just write down the factorization of the polynomial, I think.

4. May 20, 2012

### Staff: Mentor

Given: aα²+bα + c = 0 .................... (i)
it factorizes: aα(α + b/a) + c = 0

Substitute for the bold α.

5. May 20, 2012

### physics kiddy

Thank you very very very very much. I can't explain how happy I am to get the answer. Thanks again.

6. May 20, 2012

### SammyS

Staff Emeritus
Yes, that's correct supposing that physics kiddy knows:
If α and β are the solutions to $\text{a}x^2+bx+c=0\,,$ then $\text{a}x^2+bx+c=\text{a}(x-\alpha)(x-\beta)\ .$​

Of course, I agree that is a very handy thing to know!